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Archive 0 - 2005 October 15 Saturday |
Kaimbridge's Comment/Discussion Board
Hmm...very interesting question...What is the context of the original problem? -- HappyCamper 01:49, 30 October 2005 (UTC)
Hi Kaimbridge. I have one remark. According to some style rule, one should not put links in section headings, rather below in the text. For this reason I undid your change at quotient rule. I understand that sometimes links in section headings are more straightforward, but people say they don't look good there. :) (and I agree) Oleg Alexandrov ( talk) 16:17, 7 December 2005 (UTC)
I delinked "Aspect ratio" becuase disambiguation pages aren't meant to hold any information. If the Template:Planet Infobox/Earth was linking to Aspect ratio to access the dictionary defintion "The aspect ratio of a two-dimensional shape is the ratio of its longest dimension to its shortest dimension." then this is a problem because Wikipedia is not dictionary and that definiton will be removed eventually. You should only create an article Aspect ratio (ellipse) is it is more than a dictionary definition - or it will be deleted. If you want to link "Aspect ratio" to a definition, then I recommend a Wiktionary link.-- Commander Keane 16:48, 10 December 2005 (UTC)
I replied on my talk page. Sorry it took me so long, it got buried in other discussion. Oleg Alexandrov ( talk) 02:08, 20 December 2005 (UTC)
Hi,
I suspected something like that, but I wasn't sure. "mezzano" is actually derived from "mezzo", which means half, but it's a word used exclusively for pieces of forniture divided into two or more sections :-)) I'm going to reinsert your data with a fixed translation (medio (average) will do :-)) Cheers,
Alfio
17:45, 29 December 2005 (UTC)
(Moved original question from Kbolino's Talk page)
Is there a way, instead, to make phi in sans-serif to look like "regular" phi? The standard symbol for latitude is regular phi (see Vincenty's formula (PDF)). ~Kaimbridge~ 19:36, 19 February 2006 (UTC)
I think the additional material you've put in the rhumb line article about the inverse Gudermannian function is great, but I disagree with your decision to replace all the previous equations with your new formulation. The other form, with logs, secants, and tangents, is much more common, and I think makes a great deal more sense to beginners, as most people who would read this article would know what the ln, sec, and tan functions are, without necessarily knowing about inverse hyperbolic trig functions. -- Dantheox 01:32, 23 February 2006 (UTC)
Hello Kaimbridge, thanks for the feedback. I'm sorry, though, I can't really help you out. I'm just starting to learn about the subject, and made the page primarily for disambiguation purposes. I was going to refer you to a page, but it turned out it was one you wrote, so I don't think it would have been any help ^_^ Also, most of the other pages I've been reading only deal with circular arcs, so they're no help either. Again, sorry I couldn't be more help. -- Nekura 16:22, 9 June 2006 (UTC)
If you must write "antiln" instead of the standard "exp", I think your TeX code should say \opperatorname{antiln}, instead of the more complicated sequence you used. Is there a reason to prefer "antiln" to "exp"? Michael Hardy 21:57, 9 June 2006 (UTC)
Haha, my lack of knowledge is showing up. I was working on Wikipedia:WikiProject Missing encyclopedic articles and 'oblateness constant' showed up. I attempted to define it on oblate and redirected the original, but it could do with a professional's view! Please do change it. Cheers akaDruid 15:26, 23 June 2006 (UTC)
Done :)
You can do whatever you want with the article now, I'm done with it. -- Michael Retriever 17:16, 3 August 2006 (UTC)
Dear Kaimbridge,
The means available in the Earth radius article weren't satisfactory enough for me, that's why I searched an ideal Earth radius given that its true equation were x^2/a^2 + y^2/a^2 + z^2/b^2 = 1. Nevertheless, I suspected that the result would be equal or nearly equal to the quadratic mean, as I must say I was quite impressed to see that there is such a powerful yet simple averaging technique for three-dimensional cartesian spaces (I didn't know it before :P ).
If it's not, then my radius mean and the quadratic mean are incorrect.
About the z matter and other possible functions for ellipse-related matters, I'm not fond of trigonometric functions. That explains why I used that function instead of the ones you propose :) What I think is important to mention however is that the function you used to calculate the perimeter/longitude of the Earth's great ellipse in the Meridional Earth radius is an approximation, and not the exact precise function (I think you already know this :P ). The true integral function is
this one; or the integral
you showed yourself?, which is a rather nicer one actually.
Well, the thing is that for what I know, all integral functions which cannot be integrated with regular techniques can be translated into infinite polynomial summations (that's what the elonged S symbol in integrals means after all), and that's what I just used in my calculus. The z function I used is here, but there are many other summations which can be used in an equally powerful manner, take for example Euler's early one.
a^2 = (n+1)b^2; (a^2/b^2) - 1 = n; n = 0.0067395962783419227214499918750432
After that, you calculate the infinite series (I did it until - (273922023375/106542032486400) (n^8) = 1.0944023775784746584284715852771e-20 using the Microsoft Calculator :P ), multiply it with 4(1/2)(π)(b) = (2)(π)(b), and you get that the Earth's great ellipse longitude is 40007861.930850134920636564989507. I did the same with z (until + (654729075/3715891200)^2 · e^20/19 = 2.9540033942026385623871168669082e-25) and I got that the ellipse longitude is 40007861.930850134920638871082041. So as you can see, calculating the exact length of an ellipse's longitude is possible, it all depends on how many terms you use in the infinite summations. Muir's and Ramanujan's approximations are, after all, fairly precise approximations.
There is also another matter which I want to discuss, about your Meridional Earth radius mean. When you calculate this mean, you're making the average radius of the Earth where (longitude of the great ellipse)/(2π)= Mr as I understand it. But this is of no use when you consider that the Earth is three-dimensional and not two-dimensional, is it? This mean is to find a radius of a supposed circular shape of the great ellipse, and not to find a radius of a supposed shperical shape of the ellipsoid. I'm not sure if I make myself clear, this is one of these things which is simpler to explain with a drawing; if you don't understand me please do tell.--
Michael Retriever
13:36, 5 August 2006 (UTC)
Right - good point. The so-called "egg" is now vertical :-) Deuar 15:16, 23 August 2006 (UTC)
Hiya Kris! Regarding "E"arth vs. "e"arth, isn't a good guideline, if it is preceeded by "the", it means Earth as a common noun, in the general, planetary sense, not requiring capitalization (e.g., "but the earth's composition" = "but the planet's composition"), whereas "Earth's..." refers to Earth as a proper noun? I've found a good way to decide is replace "Earth" with either "Mars" or "Venus": Would you say "but the Mars' composition" or "but the Venus' composition"? No, you would say either "but Mars' composition" or "but the planet Venus' composition"——right? P=) ~Kaimbridge~ 14:26, 30 August 2006 (UTC)
I've done an Excel database to solve the problem of the perimeter of an ellipse, and also a Word document explaining how to find the distance between two coordinates on Earth. You can download it at http://www.megaupload.com/?d=KAV5AZ8X. I am sure there are other more sophisticated ways to look for a solution, but I've done it according to my maths level. I'm used to cartesian coordinates, and I know very few things about differential calculus.
This solution I found does have an accuracy of millimetres. Ground elevation isn't mentioned, I believe that is elementary. Also, I want to acknowledge that, after all, I don't believe that the roEarth is perfectly spheroidal, and I think you think that also :). There probably is a more accurate Earth equation within some NASA department, describing surface irregularities, etc. I once heard that within its spheroidal shape, the Earth has a pear-shaped tendency (it is thinner at the top and fatter at the bottom). Hey but, whatever! This is what I've been able to do with the data available, and it's more than enough :P
Please do tell me if the Excel database works fine; I've never checked language compatibility before, so I don't know whether it will or it won't. I'd also be very grateful if you mention any mistake you may see within the formulas in the Word document. Thank you for the attention you've given to the matter until now :P -- Michael Retriever 19:57, 3 September 2006 (UTC)
Ok, I'll tell you what we'll do. I'd like to finish with this problem by the end of the week, so I'm turning on the boost. First, let's get on with the Earth's shape confusion. Visit Explanation of the Earth's shape.
After you've done that, tell me if you agree with it or if you don't. We discuss, find the correct equation for the Earth's shape, and I make the appropiate modifications in my files.
The data available of Mars in that webpage is different to the one given of the Earth in Wikipedia. It says:
I wish they gave us the varius equatorial radii they used for the mean, but anyway here's the best equation I can think for it:
Oh, now that's nice. Then it all comes down to:
Where p and p' are the two different equatorial radii, supossing they are the ones that correspond to x and y axes.
And about the x y z being with this radius or with the other radius, it does not matter. What is indeed important is that if you're supposed to use two radii a with two dimensional variables, you use them with two dimensional variables; and that once you choose, you are coherent with yourself.
You can use
instead of
if you like. But then, when you use a three-dimensional Cartesian coordinate (for example (1,6,3) ), you must associate each individual coordinate with its right dimensional variable ( 1 to x, 6 to y, and 3 to z ) and with its right axis in a drawing of the figure. This is easy when you only work with your own drawing and your own coordinates, but if someone who's using the second equation instead of the first one gives you a three-dimensional coordinate, and you try to look at it in your equation or in your drawing, you'll have to change the order of them. I use the setup I use because it's the one I've seen the most.
And after this... how come there is more data about Mars than about the Earth itself?
The more data for different axis they give you, the more accurate the equation is. But you cannot change its basic form because we are always using a three-dimensional Cartesian space, and what I mean by not changing its basic form is that each radius you give me must still be associated with the right dimensional variable. You can have up to an hexaxial shape, but the dimensional variables are still only three. The one for height goes with height radii, the one for width goes with width radii, and the one for depth goes with depth radii. Hee hee, and you missed the spelling there, it's quadraxial for quadrātus.
Next point: about the series expansion for the perimeter of an ellipse. What I like about the formula I use is that it is very elegant-looking for people experienced in technical drawing. When you make a building or garden plan or any drawing of mainly common geometric shapes, and you draw an ellipse in it, you mention the eccentricity of it. Everyone knows what you mean when you say "the eccentricity of this ellipse is 0.6". On the other hand, no one has a clue what you're talking about if you say "this ellipse's f' is 0.6, where f'=a-b/a+b", or "this ellipse's n is 0.6, where a^2=(n+1)b^2" (Euler's early formula). Thus, other formulas for the perimeter of an ellipse look weird, while the one that contains a simple series of divided integers and a fraction of a power of the eccentricity looks very intuitive and easy to remember.
The Excel database I made calculates up to 60 diminishing terms in the formula I use. For an ellipse with an eccentricity between 0 and 0.3, that is more than half the diminishing terms needed for a precision of 16 digits after the decimal separator; which is absurdly accurate for calculations with ellipses within the Earth. Probably other formulas are better for higher eccentricities, but for the problem we have at hand, they aren't useful enough to make an intuitive formula turn into something complicate. Nevertheless, as a curiosity, I'll check your f' formula along with Euler's one with ellipses of different eccentricities, to study their behaviour. The OE one I cannot test because I don't know what it means.
Last point: UBasic. I'm doing fine with the tools I have already: I have a powerful calculator, lots of paper and ink to draw and write, and some office tools in my computer (which by the way has no internet access). I'm not in real need for a maths programme, and you know I dislike formulas editors :P
I haven't tried UBasic yet. But since you insist so much, I'll try it. With one condition. You must download some spreadsheet-reading software, and try the Excel database I made. I think OpenOffice does that, but I'm not sure if there's OpenOffice for Windows (I think there is).
See you in a while. -- Michael Retriever 01:07, 6 September 2006 (UTC)
Now this is weird: Ellipse perimeters. For some reason, it seems that the formula I suggested is more accurate than the other two... -- Michael Retriever 11:51, 6 September 2006 (UTC)
99 End 100.BC(XP,RG):V_bc=1:For TN=1 To RG:V_bc=V_bc*(1+XP-TN)/TN:Next:Return(V_bc)
e=.6:z=1:for I=1 to 10:z=z+(2*I-1)*.BC(.5,I)^2*e^(2*I):?(2*I-1)*.bc(.5,I)^2*e^(2*I):Next:?:?"z =";z
10 Point 7
Oh so UBasic is one of them many tools for programming in Basic. I used VisualBasic for a short while (some years ago?), and I'm supposed to know Java; so if I'm not using them it's simply because I'm too lazy. Also, I'm supposed to know binomial expansions because it's in Computer Science's 1st year things-to-do list (thing which I'm supposed to be studying), but then you can call me lazy again even if I'm not supposed to be. I'll check the formulas, and I'll add the OE one. The Euler's early formula is the one I mentioned in our previous conversation ( press me pleease!) -- Michael Retriever 12:40, 7 September 2006 (UTC)
Wait... I know how to get the binomial expansions for your formula because you say the formula I suggested is equivalent to
thus
Ouch I forgot :P But I did use it correctly in the new study for the different formulas. I told you I hate formulas editors!! And specially this Wikipedia one, it's like... clumsy.
then I can check the formulas in the perimeters study I showed you. But I have no clue how you do that, since k>n in the binomial expansions. -- Michael Retriever 13:00, 7 September 2006 (UTC)
I've tested again the formulas, and the results are quite pretty. 10 terms in your formula are more powerful than 80 in mine! But the OE formula doesn't seem to work. You sure I have to multiply by cos((OE/2)^2) ? Ellipse perimeters. -- Michael Retriever 19:40, 7 September 2006 (UTC)
Sorry for answering this late, I didn't have access to the internet until now. Anyway, I believe the work is done: we've seen the Earth's equation, I found a way to find the distance between two coordinates on Earth, and you've found a damn good formula for the perimeter of an ellipse. There are only some minor details left I want to discuss:
Over all, after seeing your way of understanding ellipse-related matters, I must say I was kind of disturbed to see that trigonometric functions are really that powerful. I've always felt a stitch in my brain when thinking about them; school teachers are not a great example for rigurosity (e.g.):
(1) The equation of a circle is x^2+y^2=r^2, where x is the distance from the point to the origin in the x axis, and y is the distance from the point to the origin in the y axis. Great, interpretation of a graphic through elementary algebraic expressions. (2) The equation of a circle is (cos(α)·r)^2+(sin(α)·r)^2=r^2, where α is the angle between the x axis and the line that goes from the origin to the point. Uhm, okay, you describe an equation with two auxiliar functions, cos(α) and sin(α). And the equations for those functions are? The functions cos(α) and sin(α) are trigonometric functions. Which means? ... It means that to know the result of their equations you must use a calculator. ¬¬'
And thus, I never understood how they really work, because no one has ever cared to define them to me. Even more incredible is the fact that you get the same answer when asking some university professors! :/
But trigonometric functions are powerful and of great utility, so I'll care to study them carefully for some time, until I get to use them with ease.
It was a pleasure discussing the ellipse matter with you. You'll probably want to make some clean-up now in your discussion section! :P -- Michael Retriever 13:26, 12 September 2006 (UTC)
The polygonal paths 1 and 2 are just two more measures I made for the ellipses' perimeters, and they have nothing to do with the infinite binomial expansions. But hey, you poking my article? I'll poke yours then!
And now, another doubt. Going back to the discussion about the ellipse's perimeter page we both love so much; you see the second mathematical expression? Isn't it blacker than the rest, and also a bit blurred? Do you know how to change that? The same happens for b/a = cos α and α = cos^-1 (b/a).
No, I haven't tried them. Will I try them? Probably. Maybe next month; I'm off ellipses for a while.
Oh! and sin^-1 = arcsin, cos^-1 = arccos, tan^-1 = arctan. -- Michael Retriever 17:53, 13 September 2006 (UTC)
You checked if arcsin x =/= 1/sin x? Who knows, I wouldn't be surprised if it turned out to be equal :P
I forgot to put a smiley :) Your poking was just an excuse to mention a few things I wanted to say to you sooner or later. -- Michael Retriever 09:34, 14 September 2006 (UTC)
I think I understand now. Mind take a look? :) Binomial coefficients with real non-natural numbers -- Michael Retriever 19:09, 21 September 2006 (UTC)
~Kaimbridge~ 19:32, 23 September 2006 (UTC)
Hi Kaimbridge, wow - those are extended talks higher up on this page! Regarding planetary bodies, the equal axes are almost always the equatorial ones as you say, at least when you've got an oblate spheroid like the Earth etc. However check out (136108) 2003 EL61 which is thought to be prolate for reasons connected to its fast rotation (it's weird). Now, you mentioned that
It is really just a matter of what definitions of a,b,c, you take, but it is obviously inconsistent with the present statement a>=b>=c in ellipsoid!
I'm afraid i'm not well versed in what are the conventions with x,y,z labeling in cartography, spheroids, etc, so I can't say anything wise there, although I suspect the conventions may differ depending on the exact field. aargh! Good luck with the edit... :-) Deuar 19:02, 22 October 2006 (UTC)
Hi Kaimbridge: please remove your image from your signature. It is against WP:SIG#Images. You will see various reasons for this on that page. Thanks! — Mets501 ( talk) 21:28, 13 December 2006 (UTC)
Thanks. (Further reply on my talk page.) — Steve Summit ( talk) 01:07, 5 August 2007 (UTC)
Hi, thanks for the note - I've been behind in my watchlist for a while :p Before I comment, I need to find some time (sigh) to sit back and read your explanations carefully. Cheers, Deuar 10:15, 26 October 2007 (UTC)
Hi! It looks like you modified a page on it.wiki, replacing different letters with the same symbol . I reverted your edit. I inform you that such an action could be looked as vandalic. In case the edit was wrongly attributed, I beg your pardon for the inconvenience. You can contact me here. Thank you -- Zio illy ( talk) 22:11, 1 March 2008 (UTC)
You asked at Help talk:Displaying a formula how one requests a new feature. Since you asked in December I figured you may no longer be watching that page so I have come here to let you know that the place to go is the Mediawiki bugzilla and Wikipedia:Bug reports will explain how you do it. Asking questions on talk pages often does not get results depending on how well frequented they are. It is usually better to ask at WP:Helpdesk who will direct you to the right place if they do not know themselves. SpinningSpark 19:12, 26 April 2008 (UTC)
You may think you reverted me, but you were too late! I had already undone all of my edits.
I had always assumed that the latitude everyone uses is the geocentric latitude, which I why I thought the surface distances were wrong. Turns out, without knowing it, everything I've ever done has been geodetic. Oh well. Rracecarr ( talk) 18:15, 11 June 2008 (UTC)
Looking at the first formula for delta sigma, it appears it is only a function of delta lambda and position on the sphere. Shouldn't delta sigma also depend on delta phi? —Preceding unsigned comment added by 192.146.1.16 ( talk) 17:47, 27 October 2008 (UTC)
Hiya RJ! I changed wording in the
image and just wanted my comments to reflect that in the
Talk:Earth comments (I didn't change or add to the discussion as the archive advisory warns against). Isn't that legitimate?
~Kaimbridge~ (
talk)
18:29, 10 December 2008 (UTC)
What is original about that article? ~Kaimbridge~ ( talk) 20:38, 22 February 2009 (UTC)
Hey, I'm very sure about my correct to the great circle distance problem. I know that it appears to be correct for the edge cases of 0 and pi/2, which is why it took me 2 months to debug this problem. I realized something was wrong when I was trying to write a program to generate a circle on the surface of a sphere given a polar coordinate system and the currently posted formula didn't quite seem to work. I finally found that I only get the circle (using the great circle distance formula to determine a cutoff for the region in the circle) to render correctly with that change.
Please contact me at dsg [the product of 13717421 * 9] (at) gmail to discuss further. That is, if letters are As and numbers are Bs, AAABBBBBBBBB@gmail.com. I just don't want to be harvested by spambots.
Anyway, given the amount of frustration this error caused me, I'd really like it to be changed.
Thanks! —Preceding unsigned comment added by 67.183.110.231 ( talk) 06:42, 4 July 2010 (UTC)
Hi Kaimbridge,
I see you edited the HCS page.
I attended Hampshire Country School 1974-1979. Dr. Patey actually supported me through my stay and I remain the only person in the history of the school to attend on 100% scholarship. I am sure you remember the Patey's were that sort of folks...always quietly helping us kids.
I am also the only person to have returned as a teacher and Houseparent (96-97).
Anyway, what years did you attend.....it is such a small population that actually went to HCS and we usually know/know of each other. Are you going to the first ever all-years reunion we hope to hold in a year or two?
I am still very involved with the school and would love to walk you around the grounds if you should like to return and see the place.
As for the post, I am a Senior Intelligence Office with the Defense Intelligence Agency (DIA) and am currently on loan to National Geospatial Intelligence Agency (NGA) as the Lead for Enhanced Analytics; I developed BEAT and we currently practice the analytics at both agencies.
But the information does not need to be publicized……
Thanks Kevin Kevin.r.boland@nga.mil ( 214.28.226.229 ( talk) 12:50, 5 November 2012 (UTC))
Hi Kaimbridge, Thanks for adding the new section to the article binomial coefficient. I wanted to discuss one aspect of it that I changed and then you changed back, namely the use of many different (but related) variables. In my experience, this addition of extra variables makes notation cleaner (i.e., fewer symbols per equation) at the expense of being less comprehensible -- in particular, readers will have (as I did) to keep looking back to the top of the section to understand the relationship between r and f and so on. We could discuss this, or seek a third opinion; what do you prefer? Best, -- JBL ( talk) 21:34, 27 March 2013 (UTC)
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Don't understand this change -- did you mean it? According to Google Earth, at least, here 42°20′21″N 71°04′23″W / 42.3393°N 71.0731°W is midway along Washington Street more or less exactly between the two squares, but it looks like you changed it to 42°20′18″N 71°04′23″W / 42.3383°N 71.0731°W here, on E Newton St along one side of Franklin Square. Am I missing something? — Steve Summit ( talk) 04:01, 8 February 2014 (UTC)
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Comment/Discussion Board Archives |
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Archive 0 - 2005 October 15 Saturday |
Kaimbridge's Comment/Discussion Board
Hmm...very interesting question...What is the context of the original problem? -- HappyCamper 01:49, 30 October 2005 (UTC)
Hi Kaimbridge. I have one remark. According to some style rule, one should not put links in section headings, rather below in the text. For this reason I undid your change at quotient rule. I understand that sometimes links in section headings are more straightforward, but people say they don't look good there. :) (and I agree) Oleg Alexandrov ( talk) 16:17, 7 December 2005 (UTC)
I delinked "Aspect ratio" becuase disambiguation pages aren't meant to hold any information. If the Template:Planet Infobox/Earth was linking to Aspect ratio to access the dictionary defintion "The aspect ratio of a two-dimensional shape is the ratio of its longest dimension to its shortest dimension." then this is a problem because Wikipedia is not dictionary and that definiton will be removed eventually. You should only create an article Aspect ratio (ellipse) is it is more than a dictionary definition - or it will be deleted. If you want to link "Aspect ratio" to a definition, then I recommend a Wiktionary link.-- Commander Keane 16:48, 10 December 2005 (UTC)
I replied on my talk page. Sorry it took me so long, it got buried in other discussion. Oleg Alexandrov ( talk) 02:08, 20 December 2005 (UTC)
Hi,
I suspected something like that, but I wasn't sure. "mezzano" is actually derived from "mezzo", which means half, but it's a word used exclusively for pieces of forniture divided into two or more sections :-)) I'm going to reinsert your data with a fixed translation (medio (average) will do :-)) Cheers,
Alfio
17:45, 29 December 2005 (UTC)
(Moved original question from Kbolino's Talk page)
Is there a way, instead, to make phi in sans-serif to look like "regular" phi? The standard symbol for latitude is regular phi (see Vincenty's formula (PDF)). ~Kaimbridge~ 19:36, 19 February 2006 (UTC)
I think the additional material you've put in the rhumb line article about the inverse Gudermannian function is great, but I disagree with your decision to replace all the previous equations with your new formulation. The other form, with logs, secants, and tangents, is much more common, and I think makes a great deal more sense to beginners, as most people who would read this article would know what the ln, sec, and tan functions are, without necessarily knowing about inverse hyperbolic trig functions. -- Dantheox 01:32, 23 February 2006 (UTC)
Hello Kaimbridge, thanks for the feedback. I'm sorry, though, I can't really help you out. I'm just starting to learn about the subject, and made the page primarily for disambiguation purposes. I was going to refer you to a page, but it turned out it was one you wrote, so I don't think it would have been any help ^_^ Also, most of the other pages I've been reading only deal with circular arcs, so they're no help either. Again, sorry I couldn't be more help. -- Nekura 16:22, 9 June 2006 (UTC)
If you must write "antiln" instead of the standard "exp", I think your TeX code should say \opperatorname{antiln}, instead of the more complicated sequence you used. Is there a reason to prefer "antiln" to "exp"? Michael Hardy 21:57, 9 June 2006 (UTC)
Haha, my lack of knowledge is showing up. I was working on Wikipedia:WikiProject Missing encyclopedic articles and 'oblateness constant' showed up. I attempted to define it on oblate and redirected the original, but it could do with a professional's view! Please do change it. Cheers akaDruid 15:26, 23 June 2006 (UTC)
Done :)
You can do whatever you want with the article now, I'm done with it. -- Michael Retriever 17:16, 3 August 2006 (UTC)
Dear Kaimbridge,
The means available in the Earth radius article weren't satisfactory enough for me, that's why I searched an ideal Earth radius given that its true equation were x^2/a^2 + y^2/a^2 + z^2/b^2 = 1. Nevertheless, I suspected that the result would be equal or nearly equal to the quadratic mean, as I must say I was quite impressed to see that there is such a powerful yet simple averaging technique for three-dimensional cartesian spaces (I didn't know it before :P ).
If it's not, then my radius mean and the quadratic mean are incorrect.
About the z matter and other possible functions for ellipse-related matters, I'm not fond of trigonometric functions. That explains why I used that function instead of the ones you propose :) What I think is important to mention however is that the function you used to calculate the perimeter/longitude of the Earth's great ellipse in the Meridional Earth radius is an approximation, and not the exact precise function (I think you already know this :P ). The true integral function is
this one; or the integral
you showed yourself?, which is a rather nicer one actually.
Well, the thing is that for what I know, all integral functions which cannot be integrated with regular techniques can be translated into infinite polynomial summations (that's what the elonged S symbol in integrals means after all), and that's what I just used in my calculus. The z function I used is here, but there are many other summations which can be used in an equally powerful manner, take for example Euler's early one.
a^2 = (n+1)b^2; (a^2/b^2) - 1 = n; n = 0.0067395962783419227214499918750432
After that, you calculate the infinite series (I did it until - (273922023375/106542032486400) (n^8) = 1.0944023775784746584284715852771e-20 using the Microsoft Calculator :P ), multiply it with 4(1/2)(π)(b) = (2)(π)(b), and you get that the Earth's great ellipse longitude is 40007861.930850134920636564989507. I did the same with z (until + (654729075/3715891200)^2 · e^20/19 = 2.9540033942026385623871168669082e-25) and I got that the ellipse longitude is 40007861.930850134920638871082041. So as you can see, calculating the exact length of an ellipse's longitude is possible, it all depends on how many terms you use in the infinite summations. Muir's and Ramanujan's approximations are, after all, fairly precise approximations.
There is also another matter which I want to discuss, about your Meridional Earth radius mean. When you calculate this mean, you're making the average radius of the Earth where (longitude of the great ellipse)/(2π)= Mr as I understand it. But this is of no use when you consider that the Earth is three-dimensional and not two-dimensional, is it? This mean is to find a radius of a supposed circular shape of the great ellipse, and not to find a radius of a supposed shperical shape of the ellipsoid. I'm not sure if I make myself clear, this is one of these things which is simpler to explain with a drawing; if you don't understand me please do tell.--
Michael Retriever
13:36, 5 August 2006 (UTC)
Right - good point. The so-called "egg" is now vertical :-) Deuar 15:16, 23 August 2006 (UTC)
Hiya Kris! Regarding "E"arth vs. "e"arth, isn't a good guideline, if it is preceeded by "the", it means Earth as a common noun, in the general, planetary sense, not requiring capitalization (e.g., "but the earth's composition" = "but the planet's composition"), whereas "Earth's..." refers to Earth as a proper noun? I've found a good way to decide is replace "Earth" with either "Mars" or "Venus": Would you say "but the Mars' composition" or "but the Venus' composition"? No, you would say either "but Mars' composition" or "but the planet Venus' composition"——right? P=) ~Kaimbridge~ 14:26, 30 August 2006 (UTC)
I've done an Excel database to solve the problem of the perimeter of an ellipse, and also a Word document explaining how to find the distance between two coordinates on Earth. You can download it at http://www.megaupload.com/?d=KAV5AZ8X. I am sure there are other more sophisticated ways to look for a solution, but I've done it according to my maths level. I'm used to cartesian coordinates, and I know very few things about differential calculus.
This solution I found does have an accuracy of millimetres. Ground elevation isn't mentioned, I believe that is elementary. Also, I want to acknowledge that, after all, I don't believe that the roEarth is perfectly spheroidal, and I think you think that also :). There probably is a more accurate Earth equation within some NASA department, describing surface irregularities, etc. I once heard that within its spheroidal shape, the Earth has a pear-shaped tendency (it is thinner at the top and fatter at the bottom). Hey but, whatever! This is what I've been able to do with the data available, and it's more than enough :P
Please do tell me if the Excel database works fine; I've never checked language compatibility before, so I don't know whether it will or it won't. I'd also be very grateful if you mention any mistake you may see within the formulas in the Word document. Thank you for the attention you've given to the matter until now :P -- Michael Retriever 19:57, 3 September 2006 (UTC)
Ok, I'll tell you what we'll do. I'd like to finish with this problem by the end of the week, so I'm turning on the boost. First, let's get on with the Earth's shape confusion. Visit Explanation of the Earth's shape.
After you've done that, tell me if you agree with it or if you don't. We discuss, find the correct equation for the Earth's shape, and I make the appropiate modifications in my files.
The data available of Mars in that webpage is different to the one given of the Earth in Wikipedia. It says:
I wish they gave us the varius equatorial radii they used for the mean, but anyway here's the best equation I can think for it:
Oh, now that's nice. Then it all comes down to:
Where p and p' are the two different equatorial radii, supossing they are the ones that correspond to x and y axes.
And about the x y z being with this radius or with the other radius, it does not matter. What is indeed important is that if you're supposed to use two radii a with two dimensional variables, you use them with two dimensional variables; and that once you choose, you are coherent with yourself.
You can use
instead of
if you like. But then, when you use a three-dimensional Cartesian coordinate (for example (1,6,3) ), you must associate each individual coordinate with its right dimensional variable ( 1 to x, 6 to y, and 3 to z ) and with its right axis in a drawing of the figure. This is easy when you only work with your own drawing and your own coordinates, but if someone who's using the second equation instead of the first one gives you a three-dimensional coordinate, and you try to look at it in your equation or in your drawing, you'll have to change the order of them. I use the setup I use because it's the one I've seen the most.
And after this... how come there is more data about Mars than about the Earth itself?
The more data for different axis they give you, the more accurate the equation is. But you cannot change its basic form because we are always using a three-dimensional Cartesian space, and what I mean by not changing its basic form is that each radius you give me must still be associated with the right dimensional variable. You can have up to an hexaxial shape, but the dimensional variables are still only three. The one for height goes with height radii, the one for width goes with width radii, and the one for depth goes with depth radii. Hee hee, and you missed the spelling there, it's quadraxial for quadrātus.
Next point: about the series expansion for the perimeter of an ellipse. What I like about the formula I use is that it is very elegant-looking for people experienced in technical drawing. When you make a building or garden plan or any drawing of mainly common geometric shapes, and you draw an ellipse in it, you mention the eccentricity of it. Everyone knows what you mean when you say "the eccentricity of this ellipse is 0.6". On the other hand, no one has a clue what you're talking about if you say "this ellipse's f' is 0.6, where f'=a-b/a+b", or "this ellipse's n is 0.6, where a^2=(n+1)b^2" (Euler's early formula). Thus, other formulas for the perimeter of an ellipse look weird, while the one that contains a simple series of divided integers and a fraction of a power of the eccentricity looks very intuitive and easy to remember.
The Excel database I made calculates up to 60 diminishing terms in the formula I use. For an ellipse with an eccentricity between 0 and 0.3, that is more than half the diminishing terms needed for a precision of 16 digits after the decimal separator; which is absurdly accurate for calculations with ellipses within the Earth. Probably other formulas are better for higher eccentricities, but for the problem we have at hand, they aren't useful enough to make an intuitive formula turn into something complicate. Nevertheless, as a curiosity, I'll check your f' formula along with Euler's one with ellipses of different eccentricities, to study their behaviour. The OE one I cannot test because I don't know what it means.
Last point: UBasic. I'm doing fine with the tools I have already: I have a powerful calculator, lots of paper and ink to draw and write, and some office tools in my computer (which by the way has no internet access). I'm not in real need for a maths programme, and you know I dislike formulas editors :P
I haven't tried UBasic yet. But since you insist so much, I'll try it. With one condition. You must download some spreadsheet-reading software, and try the Excel database I made. I think OpenOffice does that, but I'm not sure if there's OpenOffice for Windows (I think there is).
See you in a while. -- Michael Retriever 01:07, 6 September 2006 (UTC)
Now this is weird: Ellipse perimeters. For some reason, it seems that the formula I suggested is more accurate than the other two... -- Michael Retriever 11:51, 6 September 2006 (UTC)
99 End 100.BC(XP,RG):V_bc=1:For TN=1 To RG:V_bc=V_bc*(1+XP-TN)/TN:Next:Return(V_bc)
e=.6:z=1:for I=1 to 10:z=z+(2*I-1)*.BC(.5,I)^2*e^(2*I):?(2*I-1)*.bc(.5,I)^2*e^(2*I):Next:?:?"z =";z
10 Point 7
Oh so UBasic is one of them many tools for programming in Basic. I used VisualBasic for a short while (some years ago?), and I'm supposed to know Java; so if I'm not using them it's simply because I'm too lazy. Also, I'm supposed to know binomial expansions because it's in Computer Science's 1st year things-to-do list (thing which I'm supposed to be studying), but then you can call me lazy again even if I'm not supposed to be. I'll check the formulas, and I'll add the OE one. The Euler's early formula is the one I mentioned in our previous conversation ( press me pleease!) -- Michael Retriever 12:40, 7 September 2006 (UTC)
Wait... I know how to get the binomial expansions for your formula because you say the formula I suggested is equivalent to
thus
Ouch I forgot :P But I did use it correctly in the new study for the different formulas. I told you I hate formulas editors!! And specially this Wikipedia one, it's like... clumsy.
then I can check the formulas in the perimeters study I showed you. But I have no clue how you do that, since k>n in the binomial expansions. -- Michael Retriever 13:00, 7 September 2006 (UTC)
I've tested again the formulas, and the results are quite pretty. 10 terms in your formula are more powerful than 80 in mine! But the OE formula doesn't seem to work. You sure I have to multiply by cos((OE/2)^2) ? Ellipse perimeters. -- Michael Retriever 19:40, 7 September 2006 (UTC)
Sorry for answering this late, I didn't have access to the internet until now. Anyway, I believe the work is done: we've seen the Earth's equation, I found a way to find the distance between two coordinates on Earth, and you've found a damn good formula for the perimeter of an ellipse. There are only some minor details left I want to discuss:
Over all, after seeing your way of understanding ellipse-related matters, I must say I was kind of disturbed to see that trigonometric functions are really that powerful. I've always felt a stitch in my brain when thinking about them; school teachers are not a great example for rigurosity (e.g.):
(1) The equation of a circle is x^2+y^2=r^2, where x is the distance from the point to the origin in the x axis, and y is the distance from the point to the origin in the y axis. Great, interpretation of a graphic through elementary algebraic expressions. (2) The equation of a circle is (cos(α)·r)^2+(sin(α)·r)^2=r^2, where α is the angle between the x axis and the line that goes from the origin to the point. Uhm, okay, you describe an equation with two auxiliar functions, cos(α) and sin(α). And the equations for those functions are? The functions cos(α) and sin(α) are trigonometric functions. Which means? ... It means that to know the result of their equations you must use a calculator. ¬¬'
And thus, I never understood how they really work, because no one has ever cared to define them to me. Even more incredible is the fact that you get the same answer when asking some university professors! :/
But trigonometric functions are powerful and of great utility, so I'll care to study them carefully for some time, until I get to use them with ease.
It was a pleasure discussing the ellipse matter with you. You'll probably want to make some clean-up now in your discussion section! :P -- Michael Retriever 13:26, 12 September 2006 (UTC)
The polygonal paths 1 and 2 are just two more measures I made for the ellipses' perimeters, and they have nothing to do with the infinite binomial expansions. But hey, you poking my article? I'll poke yours then!
And now, another doubt. Going back to the discussion about the ellipse's perimeter page we both love so much; you see the second mathematical expression? Isn't it blacker than the rest, and also a bit blurred? Do you know how to change that? The same happens for b/a = cos α and α = cos^-1 (b/a).
No, I haven't tried them. Will I try them? Probably. Maybe next month; I'm off ellipses for a while.
Oh! and sin^-1 = arcsin, cos^-1 = arccos, tan^-1 = arctan. -- Michael Retriever 17:53, 13 September 2006 (UTC)
You checked if arcsin x =/= 1/sin x? Who knows, I wouldn't be surprised if it turned out to be equal :P
I forgot to put a smiley :) Your poking was just an excuse to mention a few things I wanted to say to you sooner or later. -- Michael Retriever 09:34, 14 September 2006 (UTC)
I think I understand now. Mind take a look? :) Binomial coefficients with real non-natural numbers -- Michael Retriever 19:09, 21 September 2006 (UTC)
~Kaimbridge~ 19:32, 23 September 2006 (UTC)
Hi Kaimbridge, wow - those are extended talks higher up on this page! Regarding planetary bodies, the equal axes are almost always the equatorial ones as you say, at least when you've got an oblate spheroid like the Earth etc. However check out (136108) 2003 EL61 which is thought to be prolate for reasons connected to its fast rotation (it's weird). Now, you mentioned that
It is really just a matter of what definitions of a,b,c, you take, but it is obviously inconsistent with the present statement a>=b>=c in ellipsoid!
I'm afraid i'm not well versed in what are the conventions with x,y,z labeling in cartography, spheroids, etc, so I can't say anything wise there, although I suspect the conventions may differ depending on the exact field. aargh! Good luck with the edit... :-) Deuar 19:02, 22 October 2006 (UTC)
Hi Kaimbridge: please remove your image from your signature. It is against WP:SIG#Images. You will see various reasons for this on that page. Thanks! — Mets501 ( talk) 21:28, 13 December 2006 (UTC)
Thanks. (Further reply on my talk page.) — Steve Summit ( talk) 01:07, 5 August 2007 (UTC)
Hi, thanks for the note - I've been behind in my watchlist for a while :p Before I comment, I need to find some time (sigh) to sit back and read your explanations carefully. Cheers, Deuar 10:15, 26 October 2007 (UTC)
Hi! It looks like you modified a page on it.wiki, replacing different letters with the same symbol . I reverted your edit. I inform you that such an action could be looked as vandalic. In case the edit was wrongly attributed, I beg your pardon for the inconvenience. You can contact me here. Thank you -- Zio illy ( talk) 22:11, 1 March 2008 (UTC)
You asked at Help talk:Displaying a formula how one requests a new feature. Since you asked in December I figured you may no longer be watching that page so I have come here to let you know that the place to go is the Mediawiki bugzilla and Wikipedia:Bug reports will explain how you do it. Asking questions on talk pages often does not get results depending on how well frequented they are. It is usually better to ask at WP:Helpdesk who will direct you to the right place if they do not know themselves. SpinningSpark 19:12, 26 April 2008 (UTC)
You may think you reverted me, but you were too late! I had already undone all of my edits.
I had always assumed that the latitude everyone uses is the geocentric latitude, which I why I thought the surface distances were wrong. Turns out, without knowing it, everything I've ever done has been geodetic. Oh well. Rracecarr ( talk) 18:15, 11 June 2008 (UTC)
Looking at the first formula for delta sigma, it appears it is only a function of delta lambda and position on the sphere. Shouldn't delta sigma also depend on delta phi? —Preceding unsigned comment added by 192.146.1.16 ( talk) 17:47, 27 October 2008 (UTC)
Hiya RJ! I changed wording in the
image and just wanted my comments to reflect that in the
Talk:Earth comments (I didn't change or add to the discussion as the archive advisory warns against). Isn't that legitimate?
~Kaimbridge~ (
talk)
18:29, 10 December 2008 (UTC)
What is original about that article? ~Kaimbridge~ ( talk) 20:38, 22 February 2009 (UTC)
Hey, I'm very sure about my correct to the great circle distance problem. I know that it appears to be correct for the edge cases of 0 and pi/2, which is why it took me 2 months to debug this problem. I realized something was wrong when I was trying to write a program to generate a circle on the surface of a sphere given a polar coordinate system and the currently posted formula didn't quite seem to work. I finally found that I only get the circle (using the great circle distance formula to determine a cutoff for the region in the circle) to render correctly with that change.
Please contact me at dsg [the product of 13717421 * 9] (at) gmail to discuss further. That is, if letters are As and numbers are Bs, AAABBBBBBBBB@gmail.com. I just don't want to be harvested by spambots.
Anyway, given the amount of frustration this error caused me, I'd really like it to be changed.
Thanks! —Preceding unsigned comment added by 67.183.110.231 ( talk) 06:42, 4 July 2010 (UTC)
Hi Kaimbridge,
I see you edited the HCS page.
I attended Hampshire Country School 1974-1979. Dr. Patey actually supported me through my stay and I remain the only person in the history of the school to attend on 100% scholarship. I am sure you remember the Patey's were that sort of folks...always quietly helping us kids.
I am also the only person to have returned as a teacher and Houseparent (96-97).
Anyway, what years did you attend.....it is such a small population that actually went to HCS and we usually know/know of each other. Are you going to the first ever all-years reunion we hope to hold in a year or two?
I am still very involved with the school and would love to walk you around the grounds if you should like to return and see the place.
As for the post, I am a Senior Intelligence Office with the Defense Intelligence Agency (DIA) and am currently on loan to National Geospatial Intelligence Agency (NGA) as the Lead for Enhanced Analytics; I developed BEAT and we currently practice the analytics at both agencies.
But the information does not need to be publicized……
Thanks Kevin Kevin.r.boland@nga.mil ( 214.28.226.229 ( talk) 12:50, 5 November 2012 (UTC))
Hi Kaimbridge, Thanks for adding the new section to the article binomial coefficient. I wanted to discuss one aspect of it that I changed and then you changed back, namely the use of many different (but related) variables. In my experience, this addition of extra variables makes notation cleaner (i.e., fewer symbols per equation) at the expense of being less comprehensible -- in particular, readers will have (as I did) to keep looking back to the top of the section to understand the relationship between r and f and so on. We could discuss this, or seek a third opinion; what do you prefer? Best, -- JBL ( talk) 21:34, 27 March 2013 (UTC)
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Don't understand this change -- did you mean it? According to Google Earth, at least, here 42°20′21″N 71°04′23″W / 42.3393°N 71.0731°W is midway along Washington Street more or less exactly between the two squares, but it looks like you changed it to 42°20′18″N 71°04′23″W / 42.3383°N 71.0731°W here, on E Newton St along one side of Franklin Square. Am I missing something? — Steve Summit ( talk) 04:01, 8 February 2014 (UTC)
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